Let \({E}\)  be associated with \({C}\mathrel {\mathop :}={B}(X) \cap \lbrace  A :\mu (A) < \infty \rbrace \)  in Lemma REF . Recall REF 
that \({C}\)  is \({N}_\mu \) -generating. If \({E}\)  is countable,
then \(\bigcup {E}\)  is measurable and, accordingly, an
\({N}_\mu \) -essential upper bound of \({E}\) . Thus \(X \setminus \bigcup {E}\in {N}_\mu \) , since \({E}\)  is \({N}_\mu \) -generating. We
have proven that \(\mu \)  is \(\sigma \) -finite.
On the other hand, if \({E}\)  is uncountable, the Continuum
Hypothesis guarantees that it has cardinal greater or equal to
\(\mathfrak {c}\) . Assume if possible that \((X, {B}(X), \mu )\)  is
localizable. As the map \({B}(X) \rightarrow {B}(X) / {N}_{\mu }\)  is
onto, we deduce from Lemma REF  that \(\operatorname{\mathrm {card}}{B}(X) \geqslant 2^{\mathfrak {c}} > \mathfrak {c}\) . However, Borel sets are Suslin,
and Suslin sets are continuous images of closed subsets of a
particular Polish space, the Baire space, see e.g
[1]}. This gives the upper bound \(\operatorname{\mathrm {card}}{B}(X)\leqslant \mathfrak {c}\) , contradicting the preceding inequality.
